Let $$\alpha,\beta,\gamma$$ and $$\delta$$ be the coefficients of $$x^{7},x^{5},x^{3}$$ and x respectively in the expansion of $$(x+\sqrt{x^{3}-1})^{5}+(x-\sqrt{x^{3}-1})^{5},x > 1$$.If u and v satisfy the equations $$\alpha u+\beta v=18\\ \gamma u+\delta v=20$$ then u+v equals:

The expression $$(x + \sqrt{x^3 - 1})^5 + (x - \sqrt{x^3 - 1})^5$$ for $$x>1$$ simplifies by letting $$a = x$$ and $$b = \sqrt{x^3 - 1}$$, so it becomes $$(a + b)^5 + (a - b)^5$$. Since the binomial theorem gives $$(a + b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5a b^4 + b^5$$ and $$(a - b)^5 = a^5 - 5a^4b + 10a^3b^2 - 10a^2b^3 + 5a b^4 - b^5$$, adding these expansions eliminates all odd powers of $$b$$.

This gives $$(a + b)^5 + (a - b)^5 = 2a^5 + 20a^3b^2 + 10ab^4 = 2\bigl(a^5 + 10a^3b^2 + 5ab^4\bigr)\,. $$

Substituting back $$a = x$$ and $$b^2 = x^3 - 1$$ (so that $$b^4 = (x^3 - 1)^2 = x^6 - 2x^3 + 1$$) transforms the expression into $$2\bigl[x^5 + 10x^3(x^3 - 1) + 5x(x^6 - 2x^3 + 1)\bigr]\,. $$

Expanding inside the brackets yields $$x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x = 5x^7 + 10x^6 + x^5 - 10x^4 - 10x^3 + 5x$$, and multiplying by 2 gives $$10x^7 + 20x^6 + 2x^5 - 20x^4 - 20x^3 + 10x\,. $$

From this polynomial the coefficients are $$\alpha=10$$ for $$x^7$$, $$\beta=2$$ for $$x^5$$, $$\gamma=-20$$ for $$x^3$$, and $$\delta=10$$ for $$x$$. The conditions $$\alpha u + \beta v = 18$$ and $$\gamma u + \delta v = 20$$ thus become $$10u + 2v = 18$$ and $$-20u + 10v = 20$$. Dividing the first by 2 gives $$5u + v = 9$$, and dividing the second by 10 gives $$-2u + v = 2$$. Subtracting the second from the first yields $$7u = 7$$, so $$u = 1$$, and substituting back gives $$v = 4$$. Therefore $$u + v = 5$$.

The correct option is A. 5.